$A$ stone is allowed to fall from the top of a tower $100 \, m$ high and at the same time another stone is projected vertically upwards from the ground with a velocity of $25 \, m s^{-1}$. Calculate when and where the two stones will meet? (Take $g = 10 \, m s^{-2}$).

(A) Let stone $A$ fall from the top of a tower of height $h = 100 \, m$. Its initial velocity $u_1 = 0$ and acceleration $a_1 = +g = 10 \, m s^{-2}$.
Another stone $B$ is projected upwards from the ground with initial velocity $u_2 = 25 \, m s^{-1}$ and acceleration $a_2 = -g = -10 \, m s^{-2}$.
Let the two stones meet at a point $C$ at a distance $y$ below the top of the tower after time $t$. For stone $A$:
$y = u_1 t + \frac{1}{2} a_1 t^2 = 0 + \frac{1}{2} \times 10 \times t^2 = 5t^2$ ... $(1)$
For stone $B$,the distance covered from the ground is $(100 - y)$:
$100 - y = u_2 t + \frac{1}{2} a_2 t^2 = 25t - \frac{1}{2} \times 10 \times t^2 = 25t - 5t^2$ ... $(2)$
Adding equations $(1)$ and $(2)$:
$y + (100 - y) = 5t^2 + 25t - 5t^2$
$100 = 25t$
$t = 4 \, s$
Substituting $t = 4 \, s$ into equation $(1)$:
$y = 5 \times (4)^2 = 5 \times 16 = 80 \, m$.
The stones will meet after $4 \, s$ at a distance of $80 \, m$ from the top of the tower (or $20 \, m$ from the ground).